Algebra Problems

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Age Problem –
GivenJason is 4 times older than Bob at present. 8 years ago Jason was 12 times older. How old are Jason & Bob?

Solutionx for Bobs age and 4x for Jason So 4x = x But 8 years ago Jason was 12 times older so 4x - 8 = 12(x-8) 4x - 8 = 12x - 96 88 = 8x x = 11 So Bob is currently 11 years old and Jason is 4(11) or 44 years old.

GivenBob is currently twice as old as Steve. Twenty years ago Bob was 6 times as old as Steve. What are their current ages?

To find the time that they are travelling. If the return trip took 4 hours less time. so x = 25 km/h Time for going = 350/25 = 14 h time for returning = 350/(25+10)= 10 h
GivenTom's snowmobile travels 8km/h faster than Amy's. 25 we consider the positive figure.875 =0 factorize this equation.
. Amy's speed is X and distance is 360km while Tom's speed is X+8 and distance is 400km. You can use either set of numbers. mr. (I tend to use the easier of the 2!--Amy's) 360 = X * 5 Divide by 5 X = 72 So.GivenOn a round trip of 350 km in each direction. (x+35)(x-25)=0 x = -35. Amy was going 72 km/h and Tom was going 80 km/h (72+8). Tom can travel 400. In the time that it takes Amy to travel 360km. Find the speed of each snowmobile. 350x = (350-4x)(x+10) multiply and simplify gives youx² + 10x . you first need to determine the difference in their distance and their rates.
SolutionDistance=rate * time is the equation that you are using. find the time for each direction. Plug those numbers into the equation 40 = 8*t (t=time) Divide by 8 t=5 So.
Solution350/(x+10) = (350/x)-4 350/(x+10) = (350-4x)/x cross multiply. along with the figured time. 400-360 = 40km = distance and there is an 8km/h difference in rate. Sinclair averaged 1o km/h more returning than he did going. the time that they travelled is 5 hours. to solve for X.

the digits are reversed.u) = 99 * 4 h-u=4 h=u+4 u + 4 = 2u + 1 u=3 t=0 h=7 Number is 703 Reverse is 307 703 . the number obtained is 396 less than the original number. Find the original number.100 u .Digit Problem –
GivenIf 18 is added to a two-digit number.307 = 396
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Solutionh = 2u + 1 u=t+3 396 = 100 h + 10t + u . What is the original number?
Solutionlet the no be 10x+y then 10x+y+18=10y+x then 9y-9x=18 y-x=2 x+y=8 then x=3 y=5 no. If the order of the digits is reversed. is 35
GivenThe hundred's digit of a three-digit number is one more than twice the units' digit. The units' digit is three more than the tens' digit.h = 99 (h .10t . The sum of the digits is 8.

Its speed is (B + C) mph.C = 3 [2] We have a system of equations. the boat moves AGAINST the current. [1] and [2]. It took 2 hours to go 18 miles. We have: (B + C) •~ 2 = 18 •¨ B + C = 9 [1] Going upstream. the boat moves WITH the current. Going downstream.C) mph. Add the two equations: 2B = 12 •¨ B = 6 The boat's speed in 6 mph (in still water) and the current's speed is 3 mph.
.Rate ProblemGivenA boat travels 18 miles downstream in 2 hours. What is the rate of the boat in still water and what is the rate of the current?
SolutionLet B = speed of the boat (in miles per hour). Its speed is (B . It requires 6 hours to travel back to its original starting point upstream.C) •~ 6 = 18 •¨ B . Let C = speed of the current. We have: (B . It took 6 hours to go 18 miles.

50 = $98 now 40% + 60% = 100% so $98 + $147= $245
. Check it to be safe 10 % of $245 is $24. The price has been discounted by 40 percent. What was the percent of discount?
Solution$245 : 100 percent as $147 : what percent? so (147 x100%)/245 = 14700/245 = 60 Therefore $147 is 60% of $245. The new price is 60 percent of the original price. The discounted price of the coat was $147.50 so 40% is 4 x $24.Percentage Problem –
GivenThe regular price of a coat was $245.